Solution 8.7a - Revision history

Ian at 12:14, 18 April 2010

Ian — Sun, 18 Apr 2010 12:14:04 GMT

←Older revision		Revision as of 12:14, 18 April 2010
Line 3:		Line 3:
	We calculate the position vector using <math> \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math>.		We calculate the position vector using <math> \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math>.

-	Here <math>\mathbf{u}=8\mathbf{i}+10\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>, <math>\ \mathbf{a}=-10\mathbf{j}\ </math> and <math>\ \mathbf{r}_{0}=0</math> giving,	+	Here <math>\mathbf{u}=8\mathbf{i}+10\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>, <math>\ \mathbf{a}=-10\mathbf{j}\ \text{ m}{{\text{s}}^{\text{-2}}}</math> and <math>\ \mathbf{r}_{0}=0</math> giving,

	<math> \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{j}){{t}^{\ 2}} </math>		<math> \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{j}){{t}^{\ 2}} </math>

Ian at 11:39, 18 April 2010

Ian — Sun, 18 Apr 2010 11:39:01 GMT

←Older revision		Revision as of 11:39, 18 April 2010
Line 3:		Line 3:
	We calculate the position vector using <math> \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math>.		We calculate the position vector using <math> \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math>.

-	Here <math>\mathbf{u}=8\mathbf{i}+10\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>, <math>\ \mathbf{a}=-10\mathbf{i}\ </math> and <math>\ \mathbf{r}_{0}=0</math> giving,	+	Here <math>\mathbf{u}=8\mathbf{i}+10\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>, <math>\ \mathbf{a}=-10\mathbf{j}\ </math> and <math>\ \mathbf{r}_{0}=0</math> giving,

-	<math> \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{i}){{t}^{\ 2}} </math>	+	<math> \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{j}){{t}^{\ 2}} </math>

	The <math>\mathbf{j}</math> component of this position vector is		The <math>\mathbf{j}</math> component of this position vector is

Ian: New page: The ball hits the ground when the $\mathbf{j}$ component of the position vector is zero. We calculate the position vector using $\mathbf{r}=\mathbf{u}t+\frac{1}{2}\math...$

Ian — Sun, 18 Apr 2010 11:36:02 GMT

New page: The ball hits the ground when the <math>\mathbf{j}</math> component of the position vector is zero. We calculate the position vector using <math> \mathbf{r}=\mathbf{u}t+\frac{1}{2}\math...

New page

The ball hits the ground when the <math>\mathbf{j}</math> component of the position vector is zero.

We calculate the position vector using <math> \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math>.

Here <math>\mathbf{u}=8\mathbf{i}+10\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>, <math>\ \mathbf{a}=-10\mathbf{i}\ </math> and <math>\ \mathbf{r}_{0}=0</math> giving,

<math> \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{i}){{t}^{\ 2}} </math>

The <math>\mathbf{j}</math> component of this position vector is
<math>10t-5{{t}^{\ 2}}</math>.

This is zero if <math>t=2\ \text{s}</math>.

The other solution <math>t=0</math> is the trivial solution restating the fact that the ball is at the origin at the start.

Solution 8.7a - Revision history

Ian at 12:14, 18 April 2010

Ian at 11:39, 18 April 2010

Ian: New page: The ball hits the ground when the \mathbf{j} component of the position vector is zero. We calculate the position vector using \mathbf{r}=\mathbf{u}t+\frac{1}{2}\math...

Ian: New page: The ball hits the ground when the $\mathbf{j}$ component of the position vector is zero. We calculate the position vector using $\mathbf{r}=\mathbf{u}t+\frac{1}{2}\math...$